Derivative function

A derivative of a function is what gives us the information about the rate of change of any one of the variables involved in the function, suppose x in accordance with corresponding change of any other variable in the same function, let another function be y. Derivative functions are the basics for calculus. The method of differentiation is used to find derivatives. To proceed with derivatives, first the information about limits is needed.

Limits

Limits in simple terms is the point (certain value) at which the function approaches an outcome for a given input nearly. Limit is the value at which when input is very close to the limit, the output also gets closer to it.

An example would be, for the function f(x) = 2x, as the value of x gets closer to zero, the value of the output f(x) also gets closer to zero.

Now for some functions, it also depends from which side the input value x is approaching the limit value. So we generally calculate left hand limit and right hand limit both which means x approaching limit from left side and right side respectively.

Notation

Left hand and right hand limits of a function is denoted as xa–f(x) andxa+f(x) .

Derivatives of a function can be denoted in a number of ways. To find a derivative of a function y= f(x), the derivative of the function can be denoted as; 

1. df/dx

2. f’(x)

Derivation

Suppose f is any real valued function for which we are to calculate the derivatives and in that function ‘a’ is any point in its domain of definition. Then the derivative of f at the point a will be defined by

h0[f(a+h)-f(a)]/h

It should be provided beforehand that this limit exists before the derivation is attempted. Derivative of f (x) at a is denoted by f′(a). f’(a) is the change in f(x) at a with respect to x

Example 1: Find the derivative of f(x)=7 at x=0 and at x= 7

To solve this just add the values in the formula.

f’(0)= h0[f(0+h)-f(0)]/h = h0[3-3]/h = h00/h= 0

f’(7)= h0[f(7+h)-f(7)]/h = h0[7-7]/h = h00/h= 0

It should be noted that since derivatives calculate the change in function, the derivative of a constant function should be zero at every point.

Example 2: Find the derivative of the function f(x) = 2x2 + 3x  at x = 1.

To solve this just add the values in the formula →   h0[f(a+h)-f(a)]/h

f’(1) = h0[f(1+h)-f(1)]/h= h0[{2(1+h)2+ 3(1+h)}-{2(1)2+3(1)}]/h

f’(1)= h0[{2+2h2+4h+3+3h}-{2+3}]/h = h02h2+ 7h+ 5-5/h = h02h + 7 = 2(0) + 7 = 7

Important formulas

There are some formulas that we can directly apply in order to calculate the derivatives of a more complex function. 

Firstly let’s look at algebra of derivatives of a function,

Let f and g be two functions such that their derivatives are defined in a common domain. Then (i) To calculate the derivative of a sum of two functions, it is equal to the sum of the derivatives of the functions. 

It can be written as (u+v)’= u’+v’

(ii) Similarly, the derivative of difference of two functions is equated to the difference of the derivatives of the functions. 

It can also be denoted like (u-v)’=u’-v’

(iii) Derivative of product of two functions is given by the following formula which is called product rule in differentiation. 

Another form: (uv)’=u’v+uv’

(iv) Derivative of two functions getting divided is given by the following so called quotient rule (note that in this rule, the denominator is always non–zero).

Generally written as, (u/v)’ = (u’v-uv’)/v2

Another very important results which is used often is

1. Derivative of f(x) = xn is nxn-1 for any positive integer n.

2. Let f(x) = nanxn  +   (n+1)an-1xx-2   +…+    2a2x  +  a1   be a polynomial function, where ai’s are all real numbers and an ≠ 0. Then, the derivative function is given by

df(x)/dx = nanxn-1   +    (n-1)an-1xx-2 + … + 2a2x + a1.

Examples

Using these formulas, we can solve a number of problems.

Example 1: find derivative of f(x)= (2x+1)/x2 .

We can see that this function is defined everywhere except at x=0.

Using the quotient rule here, let u = 2x+1 and v= x2.

Hence, u’= 2 and v’= 2x 

Therefore, df(x)/dx = d((2x+1)/x2)/dx = d(u/v)/dx = (u’v-uv’)/v2 = [(2)(x2) – (2x+1)(2x)]/(x2)2

= (2x2– 4x2-2x)/x4 = (-2x2-2x)/x4 = -2(1/x2 +1/x3)

Example 2 : Find the derivative of f(x) = 1 + x + x2 + x3 +… + x25 at x = 1

We know that the derivative function for f(x) = xn is nxn-1 for any positive integer n

At x = 1 the value of this function equals 

1 + 2x + 3x2+…+25x24 =1 + 2(1) + 3(1)2 + .. . + 25(1)24 = 1 + 2 + 3 + . . . + 25 = (25)(26)/2 = 325

Applications

Calculating derivatives of a function is an important step in order to move forward with the more vast subject of calculus. On its own we use derivatives to calculate the rate of change of any one variable/ quantity in accordance with another variable in the same function, tangent and normal to a curve are also calculated by using derivatives of the function of a given curve, maximum and minimum values of functions are also calculated using derivatives of a function. 

Acceleration is the derivative of velocity which is a further derivative of distance covered is a great example of derivative function applications.

Conclusion

Derivative of a function is the rate of change of one variable of the function with respect to another variable of the same function. Limits are the values at which when input is very close to the limit, the output also gets closer to it. We use limits to calculate derivatives. We also use some important results to calculate derivatives more efficiently like the product rule and quotient rule.